On 'Bi-Convex' Tiles

Let us define a 'biconvex polygonal region' as a non-convex polygonal region that can be cut into 2 convex regions. The more interesting biconvex regions are those which can be cut into 2 convex regions, which are necessarily noncongruent. Let us call them asymmetric biconvex regions.

Question: To find how good bi-convex polygons, especially asymmetric ones, are for tiling.

Here is a bi-convex pentagon that tiles the plane. It is easily seen that infinitely many such pentagons (even asymmetric ones) can be constructed.



A biconvex Hexagonal tile. Again, it is easily seen that there are infinitely many such hexagons (indeed, asymmetric ones)



A biconvex heptagon. Infinitely many such heptagons can easily be found (including asymmetric ones)



Two biconvex octagonal tiles, both asymmetric. By attaching the unit square at various positions on the long rectangle, one can make infinitely many biconvex asymmetric octagonal tiles.



Right now, I don't have any biconvex 9-gons that can tile the plane.
Take any hexagon that can tile the plane in the usual lattice. By fusing two copies of any such hexagon along an edge, we can trivially construct infinitely many biconvex 10-gons that tile the plane. However, here is an asymmetric biconvex 10-gon that tiles - it can be cut into a heptagon and a pentagon.



Questions: Are there (infinitely many) biconvex 9-gons that tile the plane? There may only be a few asymmetric biconvex 10-gons that that can tile. Which are they? And are there any biconvex 11-gons or 12-gons that can tile the plane? With higher number of tiles, the answer must be negative.